Let $p_1,\ldots,p_k$ be $k$ distinct primes (in $\mathbb{N}$) and $n>1$. Is it true that $[\mathbb{Q}(\sqrt[n]{p_1},\ldots,\sqrt[n]{p_k}):\mathbb{Q}]=n^k$? (all the roots are in $\mathbb{R}^+$)
Iurie Boreico proved here that a linear combination $\sum q_i\sqrt[n]{a_i}$ with positive rational coefficients $q_i$ (and no $\sqrt[n]{a_i}\in\mathbb{Q}$) can't be rational, but this question seems to be more difficult..

Let $\ a_i = b_i\ p_i,\ i=1,\ldots s\:,\:$ where the $p_i$ are $s$ different primes and
the $b_i$ positive integers not divisible by any of them. The author proves
by an inductive argument that, if $x_j$ are positive real roots of
$x^{n_j} - a_j = 0,\ j=1,...,s ,$ and $P(x_1,...,x_s)$ is a polynomial with
rational coefficients and of degree not greater than $n_j - 1$ with respect
to $x_j,$ then $P(x_1,...,x_s)$ can vanish only if all its coefficients vanish. $\quad$ Reviewed by W. Feller.

Let $K$ be an algebraic number field and $x_1,\ldots,x_s$ roots of the equations
$\ x_i^{n_i} = a_i\ (i=1,2,...,s)$ and suppose that (1) $K$ and all $x_i$ are real, or
(2) $K$ includes all the $n_i$ th roots of unity, i.e. $ K(x_i)$ is a Kummer field.
The following theorem is proved. A polynomial $P(x_1,...,x_s)$ with coefficients
in $K$ and of degrees in $x_i$ , less than $n_i$ for $i=1,2,\ldots s$ , can vanish only if
all its coefficients vanish, provided that the algebraic number field $K$ is such
that there exists no relation of the form $\ x_1^{m_1}\ x_2^{m_2}\:\cdots\: x_s^{m_s} = a$, where $a$ is a number in $K$ unless $\ m_i = 0\ mod\ n_i\ (i=1,2,...,s)$ . When $K$ is of the second type, the theorem was proved earlier by Hasse [Klassenkorpertheorie,
Marburg, 1933, pp. 187--195] by help of Galois groups. When K is of the first
type and K also the rational number field and the $a_i$ integers, the theorem was proved by Besicovitch in an elementary way. The author here uses a proof analogous to that used by Besicovitch [J. London Math. Soc. 15b, 3--6 (1940) these Rev. 2, 33]. $\quad$ Reviewed by H. Bergstrom.

Two nonzero real numbers are said to be equivalent with respect to a real
field $R$ if their ratio belongs to $R$ . Each real number $r \ne 0$ determines
a class $[r]$ under this equivalence relation, and these classes form a
multiplicative abelian group $G$ with identity element $[1]$. If $r_1,\cdots,r_h$
are nonzero real numbers such that $r_i^{n_i}\in R$ for some positive integers $n_i\
(i=1,...,h)$ , denote by $G(r_1,...,r_h) = G_h$ the subgroup of $G$ generated by
[r_1],...,[r_h] and by R(r_1,...,r_h) = R_h the algebraic extension field of
$R = R_0$ obtained by the adjunction of $r_1,...,r_h$ . The central problem
considered in this paper is to determine the degree and find a basis of $R_h$
over $R$ . Special cases of this problem have been considered earlier by A. S.
Besicovitch [J. London Math. Soc. 15 (1940), 3-6; MR 2, 33] and by L. J.
Mordell [Pacific J. Math. 3 (1953), 625-630; MR 15, 404]. The principal
result of this paper is the following theorem: the degree of $R_h$ with respect
to $R_{h-1}$ is equal to the index $j$ of $G_{h-1}$ in $G_h$ , and the powers $r_i^t\
(t=0,1,...,j-1)$ form a basis of $R_h$ over $R_{h-1}$ . Several interesting
applications and examples of this result are discussed. $\quad$ Reviewed by H. S. Butts

Wow, they asked Feller to review Besicovitch's paper! Well, it's not as if Feller really did much in his review, but it's funny that a probabilist would be asked to write the review.
–
KCdMay 26 '12 at 21:07

Yes, it is true that $[\mathbb{Q}(\sqrt[n]{p_1},\ldots,\sqrt[n]{p_k}):\mathbb{Q}]=n^k$ and it was proved by Besicovitch ( a student of A.A. Markov) in 1940.

Although there are (almost) infinitely many books on field and Galois theory, the only book I know which proves Besicovich's theorem (but only for odd $n$) is Roman's Field Theory (Theorem 14.3.2, page 305 of the second edition).

Yes, but the proof there is only for odd $\,n\geq 3\,$ , and the proof isn't easy at all. He mentions that the case where n is even adds no new insights and is "intricate"(!!)
–
DonAntonioMay 26 '12 at 19:44

Dear @Donantonio, you are absolutely right: thanks a lot. I suppose brave users will have to go to Richards's article which Roman follows (and mentions in his bibliography) if they want to see the intricate case of even $n$. I have made an Edit about that caveat.
–
Georges ElencwajgMay 26 '12 at 19:58

Another book that has a proof is Lisl Gaal's "Classical Galois Theory: With Examples", and she treats the case of all $n$, not just odd $n$. (In fact, one of her first steps is to reduce to the case when $n$ is even.) See Section 4.12, which essentially starts on p. 234. This is the place where I first saw the result when I was a student.
–
KCdMay 26 '12 at 21:18

A link to the section of Gaal's book where the result appears is here.
–
KCdMay 26 '12 at 21:24

Thanks for the comment and the link, @KCd. Was that the book required for your course? It doesn't seem very widespread now in American universities (but I might be mistaken: I only know about these universities through online documents [yours in particular !])
–
Georges ElencwajgMay 26 '12 at 21:42